Saved Bookmarks
| 1. |
Find the sum of magnitude of current (in Amp.) through R_(1), R_(2) and R_(3). All cells are ideal |
Answer» Solution : Potential of different points are shown. (i)current in `R_1` `I_1=(DELTAV)/R_1=(5-0)/1 A=5A` from left to right. (ii)current in `R_3` `I_3=(DeltaV)/R_3=30/1 A=30 A` from LOWER to HIGHER. (iii)For current in `R_2` using KCL `(10-x)/2+(0-x)/2+(0-x)/2+(-20-x)/1=0` `rArr 10/2-20=(3x)/2+x rArr x=-6V` `THEREFORE I_2=(20-6)/1A=14 A` |
|